Let $g(x)=x^3+12x^2+36x$ and let $c$ be the number that satisfies the Mean Value Theorem for $g$ on the interval $-8\leq x\leq-2$. What is $c$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-7$ (Choice B) B $-6$ (Choice C) C $-3$ (Choice D) D $-1$
Explanation: According to the Mean Value Theorem, there exists a number $c$ in the open interval $-8<x<-2$ such that $g'(c)$ is equal to the average rate of change of $g$ over the interval: $g'(c)=\dfrac{g(-2)-g(-8)}{(-2)-(-8)}$ First, let's find that average rate of change: $\dfrac{g(-2)-g(-8)}{(-2)-(-8)}=\dfrac{-32-(-32)}{6}={0}$ Now, let's differentiate $g$ and find the $x$ -value for which $g'(x)={0}$. $g'(x)=3x^2+24x+36$ The solutions of $g'(x)=0$ are $x=-6$ and $x=-2$. Out of these, only $x=-6$ is within the interval $-8<x<-2$. In conclusion, $c=-6$.